A 0.00400 kg bullet traveling horizontally with speed 1.00 x 103 m/s strikes a 21.1 kg door, embedding itself 10.2 cm from the side opposite the hinges as shown in the figure below. The 1.00 m wide door is free to swing on its frictionless hinges. Hinge (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Yes O No (b) If so, evaluate this angular momentum (in kg • m²/s). (If not, enter zero.) kg - m2/s If not, explain why there is no angular momentum. (c) Is mechanical energy of the bullet-door system constant in this collision? Answer without doing a calculation. Yes O No (d) At what angular speed (in rad/s) does the door swing open immediately after the collision? rad/s (e) Calculate the total energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision. (Enter your answers in (.נ KEF = KE, = (f) What If? Imagine now that the door is hanging vertically downward, hinged at the top, so that the figure is a side view of the door and bullet during the collision. What is the maximum height (in cm) that the bottom of the door will reach after the collision? cm

According to the law of conservation of momentum , the total initial momentum of the system before collision will be equal to the total final momentum of the system after collision , that is ,mbub+mdud=mb+mdv , where mb=0.004 kg is the mass of the bullet , md=21.1 kg is the mass of the door , ub=1000 m/s is the initial velocity of the bullet , ud=0 m/s is the initial velocity of the door , and v is the final velocity of the bullet-door system. The angular velocity of the bullet-door system by using the formula ω=vr , where v is the final velocity of the system and r=0.102 m is the radius of the angular motion . The kinetic energy of an object is calculated by using the formula E=12mv2 , where m is the mass and v is the velocity . According to the law of conservation of energy , energy can neither be created nor destroyed , but can only be converted from one form to another . Thus the kinetic energy after collision , will be converted to potential energy as it reaches maximum height ,that is , mgh=12mv2gh=12v2h=v22g . Calculate the final velocity of the bullet-door system by using the relation mbub+mdud=mb+mdv , where mb=0.004 kg is the mass of the bullet , md=21.1 kg is the mass of the door , ub=1000 m/s is the initial velocity of the bullet , ud=0 m/s is the initial velocity of the door , and v is the final velocity of the bullet-door system. mbub+mdud=mb+mdv0.004 kg×1000 m/s+0=0.004 kg+21.1 kgvv=4kg m/s21.104 kg≈0.2 m/sCalculate the angular velocity of the bullet-door system by using the formula ω=vr , where v≈0.2 m/s is the final velocity of the system and r=0.102 m is the radius of the angular motion . ω=vr=0.2 m/s0.102 m≈1.96 rad/sCalculate the final kinetic energy of the bullet-door system by using the formula Ef=12mb+mdv2 , where mb=0.004 kg is the mass of the bullet , md=21.1 kg is the mass of the door and v≈0.2 m/s is the velocity . Ef=12mb+mdv2=120.004 kg+21.1 kg×0.2 m/s2≈0.422 JCalculate the initial kinetic energy of the bullet by using the formula Eb=12mbub2 , where mb=0.004 kg is the mass of the bullet and ub=1000 m/s is the initial velocity of the bullet . Eb=12mbub2=12×0.004 kg×1000 m/s2=2 JCalculate the maximum height by using the relation h=v22g , where v≈0.2 m/s is the velocity of the bullet-door system and g=9.8 m/s2 is the acceleration due to gravity . h=v22g=0.2m/s22×9.8 m/s2≈2.04×10-3 m≈0.204 cmd) The angular speed is ω≈1.96 rad/s . e) The total energy of the bullet-door system , which is , Ef≈0.422 J is less than the initial velocity of the bullet , which is Eb=2 J . h) The maximum height is h≈0.204 cm .